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this question is on application of derivatives and I can't understand how the formula of the perimeter would be derived to the maximum.

Ima kun
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2 Answers2

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Let $x$ be one of the equal sides. If the angle between a side and base is $t$, we have perimeter $2x+ 2x\cos t = P \implies x = \frac{P}{2(1+\cos t)}$. Now maximize the area $\frac12(x\sin t) (2x\cos t) = x^2 \sin t \cos t =P^2 \frac{\sin t \cos t}{4(1+\cos t)^2}$. Do your own calculus and check that you will $t=\pi/3$.

anumosh
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Let the sides of the triangle be $$2x+y=P \implies 2x=P-y$$ ($y$ is the base)

Area would be:

$$\frac{y\cdot\sqrt{x^2-(\frac{y}{2})^2}}{2}$$

$$=\frac{y\sqrt{4x^2-y^2}}{4}$$

$$=\frac{y\sqrt{(P-y)^2-y^2}}{4}$$

$$=\frac{y\sqrt{P^2-2Py}}{4}$$

Now we can take derivative of this with respect to $y$ and find the maximum area. However, a trick we can do here is to square the expression $A$, find the maximum of $A^2$, and square root the result.

$$A^2=\frac{y^2(P^2-2Py)}{16}$$

$$(A^2)'=\frac{P^2y}{8}-\frac{3Py^2}{8}=0$$

$$\frac{Py}{8}(P-3y)=0$$

$$y=\frac{P}{3} \textrm{ or 0}$$

Obviously we don't one side to be $0$, so we get

$$x=y=\frac{P}{3}$$

And the triangle is an equilateral triangle!

Side note: Even if the "isoceles triangle" requirement is dropped, the optimal triangle (maximum area with fixed perimeter) is still an equilateral triangle.

Gareth Ma
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  • It would be simpler with Heron's formula... – Jean Marie May 03 '20 at 08:48
  • @JeanMarie Yes, but I do not want to prove Heron's formula here, and it's not required. This is the elementary level that I suppose is around OP's level. – Gareth Ma May 03 '20 at 08:50
  • No need to prove it ; moreover, this kind of questions often come from people accustomed to olympiads where Herons's formula is wknown. – Jean Marie May 03 '20 at 08:52
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    @JeanMarie This is nowhere near olympiad level, and calculus is generally not used in olympiad level. Heron's formula is a clear overkill here. – Gareth Ma May 03 '20 at 08:53